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It's a parabola in a uniform gravity field, an ellipse in a circular gravity field coming from a point mass. So if you want to be really pedantic, it's never a

It's a parabola in a uniform gravity field, an ellipse in a circular gravity field coming from a point mass. So if you want to be really pedantic, it's never an ellipse because the Earth is not a point mass. It would be equivalent to a point mass if the Earth were a perfect sphere of uniform density, but it isn't. In reality it's a potato like mass blob that's approximated by what geodesists call the "geoid". So in order of approximations the path of a ball thrown on earth is a parabola -> ellipse -> numerical integration of 6-dof initial conditions and spherical harmonics approximation of the earth gravity field.

centimeter 6y ago
> it's never an ellipse because the Earth is not a point mass. At least classically, a sphere is indistinguishable (gravitationally) from a point mass while you're outside it. The earth is pretty sphere-ish, locally speaking.
- libraryofbabel 6y ago
  > The earth is pretty sphere-ish, locally speaking. Kinda. There are mountains, they have their own gravitational attraction, and it can be measured... even in the 18th century! https://en.m.wikipedia.org/wiki/Schiehallion_experiment https://en.m.wikipedia.org/wiki/Schiehallion_experiment It all depends on how pedantic one wants to be :)
  - perl4ever 6y ago
    The first time I heard of the concept, it wasn't actually that, it was how in the 60s, they had to correct for mascons in the moon while orbiting.
    - garmaine 6y ago
      There are very few stable orbits close to the lunar surface. Basically a couple of polar orbits with very specific parameters, and that's it. The rest get so perturbed by gravitational anomalies that they fall out of orbit after a few months or years--faster than low Earth orbit where there is atmospheric drag! https://www.lpi.usra.edu/lunar/documents/NTRS/collection2/NASA_TP_3394.pdf https://www.lpi.usra.edu/lunar/documents/NTRS/collection2/NA...
    - Sharlin 6y ago
      Yeah, the moon's gravitational field is quite lumpy compared to Earth's, due to its smaller size and the way it is believed to have formed. Plus you can orbit much closer to the surface due to the lack of atmosphere, so the lumps are steeper.
- Tyr42 6y ago
  And a ball thrown in the air follows a parabola, locally speaking. At least, you're better off correcting for air resistance before you sub in the ellipse equation.
- Koshkin 6y ago
  Locally speaking, the earth is flat.
  - dllthomas 6y ago
    But if you're speaking globally, you've already defined yourself out of calling the earth flat.
- CydeWeys 6y ago
  Only if the sphere has uniform density. The Earth very much does not have uniform density.
avmich 6y ago
Spherical mass can be replaced with a point mass. Earth however is not spherical (biggest deviation is polar flattening). And even then, Einstein's model, unlike Newton's, says it's not an ellipse even for a point mass. So, moral of the story - we have to be not too pedantic.
- asdfasgasdgasdg 6y ago
  > Spherical mass can be replaced with a point mass I'm not a physicist, but that doesn't sound right. One example is that if you are inside of the sphere, at least some of the mass will be pulling you away from the point at its center. I think what you mean is that a point mass closely approximates a spherical mass, but the degree to which that is true becomes less and less the closer you get to the center. I don't think you have successfully contradicted what the person you responded to said.
  - klodolph 6y ago
    In Newtonian physics, a sphere and point mass are exactly interchangeable as long as you are not inside the sphere. If you are outside the sphere, the equivalence is exact, regardless of distance. Proving this is a classic problem in undergraduate physics.
    - tedunangst 6y ago
      We are inside the atmosphere at least...
      - thewakalix 6y ago
        And that brings fluid dynamics into the picture, which produces discrepancies orders of magnitude greater than the parabolic/elliptical/etc. distinction!
    - asdfasgasdgasdg 6y ago
      Thanks! Shame I never took undergraduate physics but now that you've rung my bell I think we may have discussed this in high school. What an unintuitive result that being even a meter under ground breaks what is, up to that point, a fine model.
amelius 6y ago
> an ellipse in a circular gravity field coming from a point mass. Even if you consider the point mass to move because of the mutual attraction?
- nealabq 6y ago
  One foci of the ellipse is the earth+ball center of gravity. The earth is also travelling in a much smaller ellipse around that same foci. Assuming a uniform+spherical earth. And also a uniform+spherical ball I supposed.
  - srtjstjsj 6y ago
    *One focus, of the the two foci.
pdonis 6y ago
> It's a parabola in a uniform gravity field In relativity, there is no such thing as a "uniform gravity field", if by that you mean a field where the "acceleration due to gravity" is the same everywhere. The closest you can come is the "gravity field" inside a rocket accelerating in a straight line in empty space, where the acceleration felt by the crew is constant. That kind of "gravity field" has an "acceleration due to gravity" that decreases linearly with height.
- srtjstjsj 6y ago
  Why would acceleration not be constant? (Assuming the ship isn't moving from the center of the earth outward or something like that.
  - pdonis 6y ago
    > Why would acceleration not be constant? Look up the Bell Spaceship Paradox. In relativity, two spatially separated objects (or two ends of a spatially extended single object) that have the same proper acceleration in the same direction do not stay at rest relative to each other; they move apart, as seen in each of their own frames. This is different from the behavior predicted by Newtonian mechanics. In order to have the two objects (or two ends of a single extended object) remain at rest relative to each other, the one in front has to have a smaller proper acceleration than the one in back. (Rindler coordinates are often used to describe this case.)
    - rocho 6y ago
      This is crazy, I'm finding it very hard to get an intuitive understanding of this. I guess that's why it's called "paradox".
      - pdonis 6y ago
        > I'm finding it very hard to get an intuitive understanding of this. Yes, it is certainly not intuitive. That's why Bell took the time to write a paper about it years ago. This article might help: https://www.physicsforums.com/insights/what-is-the-bell-spaceship-paradox-and-how-is-it-resolved/ https://www.physicsforums.com/insights/what-is-the-bell-spac...
kens 6y ago
> So if you want to be really pedantic, it's never an ellipse because the Earth is not a point mass. This turns out to be not pedantic but very important if you're guiding an ICBM. And when landing on the Moon, Apollo had to deal with irregularities in the Moon's gravity due to mass concentrations, called mascons. (If you're interested in missile guidance, take a look at the book Inventing Accuracy. Among other things, it discusses some of the efforts to map the Earth's gravity field to increase missile accuracy for Trident and Minuteman missiles. I knew a physicist who worked on this.)
- SiVal 6y ago
  (Ah, yes, pedantry. It has its own gravity. You can't tell me it's not a force. I can't resist its pull.) It wouldn't be an ellipse even if Earth were a point mass. The gravity of the Moon and Sun, the gravitational lumpiness in the sky, has the same effect as the gravitational lumpiness underground. The combined result may be no closer to an ellipse than it is to a parabola.
  - exDM69 6y ago
    But on low earth orbit, the L2 term of earth's oblateness dominates by an order of magnitude compared to the moon and the sun, and the rest of the planets are negligible. Source is a table in the first chapters of Fundamentals of Astrodynamics.
  - mcv 6y ago
    > "Ah, yes, pedantry. It has its own gravity. You can't tell me it's not a force. I can't resist its pull." It's not a force. From the right frame of mind[0], it's just a straight line. [0] Substituting for frame of reference.
- jhoechtl 6y ago
  To extend the pendantry: Don't we have to account that the onserved object isn't a point mass either?
- jacobwilliamroy 6y ago
  Do we have to use this knowledge to toss nukes at each other? Surely there's something else we could be doing instead?
stevespang 6y ago
. . and then you have to account for the moon's gravity in the equation depending on where it is at th3e moment you throw the ball . . .
mcv 6y ago
> "It would be equivalent to a point mass if the Earth were a perfect sphere of uniform density" Would it? Even of uniform density, the mass would not the at the core, only the center of mass would be. Most of the mass is actually closer to the surface. I'm no physicist, but I'd imagine that would have quite some impact[0] on any trajectory that crosses the surface. And for any ball thrown at human speeds, I'd expect Earth's gravity would be much closer to a uniform gravitational field than one from a point mass. [0] Pun not originally intended but I'm quite happy with it now.
TheRealDunkirk 6y ago
> So if you want to be really pedantic Based on what I've seen over the years I've been around here, I'm quite certain that's a primary motivator for a non-trivial number of commenters here.

centimeter 6y ago

> it's never an ellipse because the Earth is not a point mass. At least classically, a sphere is indistinguishable (gravitationally) from a point mass while you're outside it. The earth is pretty sphere-ish, locally speaking.

libraryofbabel 6y ago
> The earth is pretty sphere-ish, locally speaking. Kinda. There are mountains, they have their own gravitational attraction, and it can be measured... even in the 18th century! https://en.m.wikipedia.org/wiki/Schiehallion_experiment https://en.m.wikipedia.org/wiki/Schiehallion_experiment It all depends on how pedantic one wants to be :)
- perl4ever 6y ago
  The first time I heard of the concept, it wasn't actually that, it was how in the 60s, they had to correct for mascons in the moon while orbiting.
  - garmaine 6y ago
    There are very few stable orbits close to the lunar surface. Basically a couple of polar orbits with very specific parameters, and that's it. The rest get so perturbed by gravitational anomalies that they fall out of orbit after a few months or years--faster than low Earth orbit where there is atmospheric drag! https://www.lpi.usra.edu/lunar/documents/NTRS/collection2/NASA_TP_3394.pdf https://www.lpi.usra.edu/lunar/documents/NTRS/collection2/NA...
  - Sharlin 6y ago
    Yeah, the moon's gravitational field is quite lumpy compared to Earth's, due to its smaller size and the way it is believed to have formed. Plus you can orbit much closer to the surface due to the lack of atmosphere, so the lumps are steeper.
Tyr42 6y ago
And a ball thrown in the air follows a parabola, locally speaking. At least, you're better off correcting for air resistance before you sub in the ellipse equation.
Koshkin 6y ago
Locally speaking, the earth is flat.
- dllthomas 6y ago
  But if you're speaking globally, you've already defined yourself out of calling the earth flat.
CydeWeys 6y ago
Only if the sphere has uniform density. The Earth very much does not have uniform density.

avmich 6y ago

Spherical mass can be replaced with a point mass. Earth however is not spherical (biggest deviation is polar flattening). And even then, Einstein's model, unlike Newton's, says it's not an ellipse even for a point mass. So, moral of the story - we have to be not too pedantic.

asdfasgasdgasdg 6y ago
> Spherical mass can be replaced with a point mass I'm not a physicist, but that doesn't sound right. One example is that if you are inside of the sphere, at least some of the mass will be pulling you away from the point at its center. I think what you mean is that a point mass closely approximates a spherical mass, but the degree to which that is true becomes less and less the closer you get to the center. I don't think you have successfully contradicted what the person you responded to said.
- klodolph 6y ago
  In Newtonian physics, a sphere and point mass are exactly interchangeable as long as you are not inside the sphere. If you are outside the sphere, the equivalence is exact, regardless of distance. Proving this is a classic problem in undergraduate physics.
  - tedunangst 6y ago
    We are inside the atmosphere at least...
    - thewakalix 6y ago
      And that brings fluid dynamics into the picture, which produces discrepancies orders of magnitude greater than the parabolic/elliptical/etc. distinction!
  - asdfasgasdgasdg 6y ago
    Thanks! Shame I never took undergraduate physics but now that you've rung my bell I think we may have discussed this in high school. What an unintuitive result that being even a meter under ground breaks what is, up to that point, a fine model.

amelius 6y ago

> an ellipse in a circular gravity field coming from a point mass. Even if you consider the point mass to move because of the mutual attraction?

nealabq 6y ago
One foci of the ellipse is the earth+ball center of gravity. The earth is also travelling in a much smaller ellipse around that same foci. Assuming a uniform+spherical earth. And also a uniform+spherical ball I supposed.
- srtjstjsj 6y ago
  *One focus, of the the two foci.

pdonis 6y ago

> It's a parabola in a uniform gravity field In relativity, there is no such thing as a "uniform gravity field", if by that you mean a field where the "acceleration due to gravity" is the same everywhere. The closest you can come is the "gravity field" inside a rocket accelerating in a straight line in empty space, where the acceleration felt by the crew is constant. That kind of "gravity field" has an "acceleration due to gravity" that decreases linearly with height.

srtjstjsj 6y ago
Why would acceleration not be constant? (Assuming the ship isn't moving from the center of the earth outward or something like that.
- pdonis 6y ago
  > Why would acceleration not be constant? Look up the Bell Spaceship Paradox. In relativity, two spatially separated objects (or two ends of a spatially extended single object) that have the same proper acceleration in the same direction do not stay at rest relative to each other; they move apart, as seen in each of their own frames. This is different from the behavior predicted by Newtonian mechanics. In order to have the two objects (or two ends of a single extended object) remain at rest relative to each other, the one in front has to have a smaller proper acceleration than the one in back. (Rindler coordinates are often used to describe this case.)
  - rocho 6y ago
    This is crazy, I'm finding it very hard to get an intuitive understanding of this. I guess that's why it's called "paradox".
    - pdonis 6y ago
      > I'm finding it very hard to get an intuitive understanding of this. Yes, it is certainly not intuitive. That's why Bell took the time to write a paper about it years ago. This article might help: https://www.physicsforums.com/insights/what-is-the-bell-spaceship-paradox-and-how-is-it-resolved/ https://www.physicsforums.com/insights/what-is-the-bell-spac...

kens 6y ago

> So if you want to be really pedantic, it's never an ellipse because the Earth is not a point mass. This turns out to be not pedantic but very important if you're guiding an ICBM. And when landing on the Moon, Apollo had to deal with irregularities in the Moon's gravity due to mass concentrations, called mascons. (If you're interested in missile guidance, take a look at the book Inventing Accuracy. Among other things, it discusses some of the efforts to map the Earth's gravity field to increase missile accuracy for Trident and Minuteman missiles. I knew a physicist who worked on this.)

SiVal 6y ago
(Ah, yes, pedantry. It has its own gravity. You can't tell me it's not a force. I can't resist its pull.) It wouldn't be an ellipse even if Earth were a point mass. The gravity of the Moon and Sun, the gravitational lumpiness in the sky, has the same effect as the gravitational lumpiness underground. The combined result may be no closer to an ellipse than it is to a parabola.
- exDM69 6y ago
  But on low earth orbit, the L2 term of earth's oblateness dominates by an order of magnitude compared to the moon and the sun, and the rest of the planets are negligible. Source is a table in the first chapters of Fundamentals of Astrodynamics.
- mcv 6y ago
  > "Ah, yes, pedantry. It has its own gravity. You can't tell me it's not a force. I can't resist its pull." It's not a force. From the right frame of mind[0], it's just a straight line. [0] Substituting for frame of reference.
jhoechtl 6y ago
To extend the pendantry: Don't we have to account that the onserved object isn't a point mass either?
jacobwilliamroy 6y ago
Do we have to use this knowledge to toss nukes at each other? Surely there's something else we could be doing instead?

stevespang 6y ago

. . and then you have to account for the moon's gravity in the equation depending on where it is at th3e moment you throw the ball . . .

mcv 6y ago

> "It would be equivalent to a point mass if the Earth were a perfect sphere of uniform density" Would it? Even of uniform density, the mass would not the at the core, only the center of mass would be. Most of the mass is actually closer to the surface. I'm no physicist, but I'd imagine that would have quite some impact[0] on any trajectory that crosses the surface. And for any ball thrown at human speeds, I'd expect Earth's gravity would be much closer to a uniform gravitational field than one from a point mass. [0] Pun not originally intended but I'm quite happy with it now.

TheRealDunkirk 6y ago

> So if you want to be really pedantic Based on what I've seen over the years I've been around here, I'm quite certain that's a primary motivator for a non-trivial number of commenters here.